I was reading an assembly procedure that needed to align addresses on 4 bytes boundary for performance reasons so it has used the next statement that i formulated as a theorem to be proven.

Let $s$ be an integer that is not a multiple of 4 ($s \% 4 \neq 0$). $m$, the first next multiple of 4 such that $m \gt s$ is obtained using the following formula, where the operands are expressed in base 2 (Binary): $m = s + (\lnot s \land 11)$ ($\lnot a$ will return the 2's complement of $a$).

How to prove that ?

  • $\begingroup$ @dkaeae Yes. I made a mistake i forgot to tell that the negation $\lnot$ will actually return the 2's complement of its operand (The Intel's x86 neg instruction). $\endgroup$ Mar 15 '19 at 17:05
  • $\begingroup$ Yes, but it still doesn't work. Now if $s = 1$, then $m = 3$. $\endgroup$
    – dkaeae
    Mar 15 '19 at 17:07
  • $\begingroup$ In fact, $\text{and}(a, 3)$ yields a number strictly smaller than $4$ for any $a$. $\endgroup$
    – dkaeae
    Mar 15 '19 at 17:08
  • $\begingroup$ @dkaeae Damn man, i feel too dumb. That is actually the distance that is needed to reach the next multiple of 4. God, sorry for this spaghetti i made. I just corrected the formula, take a look. $\endgroup$ Mar 15 '19 at 17:43

Suppose that $s = 4a+b$, where $0 \leq b < 4$; by assumption, $b \neq 0$. If the numbers are $n$ bit long, then $\lnot s = 2^n - s = 4(2^{n-2}-a) + (4-b)$. Since $b \neq 0$, the last two bits of $\lnot s$ will be $4-b \in \{1,2,3\}$. Therefore $(\lnot s) \land (11)_2 = 4-b$. Adding this to $s$, we get $4a+b+4-b = 4(a+1)$.

  • $\begingroup$ Oh great, that's so beautifully done. I really missed the idea that $\lnot s = 2^n - s$ . That was the key for proving it. Thanks professor. The thing that puzzles me is how did the programmer knew about such a fact ?! Is it something that could be intuitively figured out and later be proved ? for example, the fact that $and(n, 3)$ always gives you the distance to the previous multiple of 4 is easy to notice and then prove. But this one is really tricky. What do you think professor @Yuval Filmus, is it a known technique for the people that work on assembly level ? $\endgroup$ Mar 15 '19 at 22:55
  • $\begingroup$ It's really impossible to say. Such tricks probably get discovered and rediscovered many times, and then spread around. You can also compute the next multiple using (s&~3)+4, something I just thought about. I'm probably not the first person figuring this one out either. $\endgroup$ Mar 15 '19 at 23:14
  • $\begingroup$ Yes, i see your point professor. Thanks again. $\endgroup$ Mar 15 '19 at 23:30

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